Question: Rewrite the equation by completing the square. $x^{2}+14x+49 = 0$ $(x + $
Solution: The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $14$, half of it is $7$, and squaring it gives us ${49}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x + 7 )^2 = 0$